Complex Conjugate Roots

You’ve seen that a quadratic with a negative discriminant has two roots like $2 + 3i$ and $2 - 3i$: a conjugate pair. That isn’t a coincidence. For any polynomial with real coefficients, non-real roots always come in conjugate pairs. This one fact lets you factor cubics and quartics from a single known root, and build polynomials with the roots you want. This page is part of AA HL (subtopic AHL 1.14).

Key ideas

How many roots?

A polynomial equation of degree $n$ has exactly $n$ roots in $\mathbb{C}$, as long as you count repeated roots as many times as they repeat. Some may be real, some non-real. The real roots are the x-intercepts of the graph; non-real roots don’t show up on the graph at all.

The conjugate root theorem

If a polynomial $p(z)$ has real coefficients and $z = a + bi$ is a root, then its conjugate $z^* = a - bi$ is also a root.

Why it works. Conjugates behave well with arithmetic: $(z + w)^* = z^* + w^*$ and $(zw)^* = z^* w^*$, so $(z^n)^* = (z^*)^n$. Real numbers are their own conjugates. So if $p(z) = a_n z^n + \cdots + a_1 z + a_0$ with real $a_k$, taking the conjugate of $p(z) = 0$ gives

$$a_n (z^*)^n + \cdots + a_1 z^* + a_0 = 0^* = 0$$

That says $p(z^*) = 0$.

Two useful consequences:

• A polynomial of odd degree with real coefficients always has at least one real root (the non-real roots pair up, so one root is left over).
• The theorem needs real coefficients. For example, $z^2 - iz + 2 = 0$ has roots $2i$ and $-i$, which are not conjugates.

The quadratic factor from a conjugate pair

If $a \pm bi$ are roots, then $(z - (a + bi))(z - (a - bi))$ is a factor. Expand it as a difference of squares:

$$\big((z - a) - bi\big)\big((z - a) + bi\big) = (z - a)^2 + b^2 = z^2 - 2az + (a^2 + b^2)$$

This is a real quadratic factor. A quick way to write it down: it’s $z^2 - (\text{sum})z + (\text{product})$, where the sum of the pair is $2a$ and the product is $a^2 + b^2$.

Using a known root to factor

To solve a cubic or quartic when you’re given one non-real root:

1. Write down its conjugate, which is also a root.
2. Build the real quadratic factor from the pair.
3. Find the other factor by dividing (polynomial division) or by comparing coefficients.
4. Solve the remaining factor.

Sum and product of roots

For $a_n z^n + a_{n-1}z^{n-1} + \cdots + a_1 z + a_0 = 0$, the roots satisfy

$$\text{sum} = \frac{-a_{n-1}}{a_n} \qquad\qquad \text{product} = \frac{(-1)^n a_0}{a_n}$$

These are covered in sum and product of roots, and they give a quick way to find a missing root or check your answer (Example 4).

Worked examples

Example 1: A quadratic from one root

One root of $z^2 + bz + c = 0$, where $b, c \in \mathbb{R}$, is $2 - 3i$. Find $b$ and $c$.

Solution. The coefficients are real, so the other root is $2 + 3i$.

• Sum of roots: $(2 - 3i) + (2 + 3i) = 4$.
• Product of roots: $(2 - 3i)(2 + 3i) = 4 + 9 = 13$.

So the quadratic is $z^2 - 4z + 13$, which means $b = -4$ and $c = 13$.

Check: the quadratic formula on $z^2 - 4z + 13 = 0$ gives $z = \dfrac{4 \pm \sqrt{-36}}{2} = 2 \pm 3i$. ✓

Example 2: Solving a cubic

Given that $1 + 2i$ is a root of $z^3 - 5z^2 + 11z - 15 = 0$, find the other roots.

Solution. The coefficients are real, so $1 - 2i$ is also a root. The quadratic factor is

$$z^2 - (\text{sum})z + (\text{product}) = z^2 - 2z + (1 + 4) = z^2 - 2z + 5$$

The cubic is this quadratic times a linear factor $(z + k)$. Compare the constant terms: $5k = -15$, so $k = -3$.

$$z^3 - 5z^2 + 11z - 15 = (z^2 - 2z + 5)(z - 3)$$

Check by expanding: $z^3 - 3z^2 - 2z^2 + 6z + 5z - 15 = z^3 - 5z^2 + 11z - 15$. ✓

The roots are $1 + 2i$, $1 - 2i$ and $3$. Their sum is $5$, matching $\dfrac{-(-5)}{1} = 5$. ✓

Example 3: Solving a quartic

Given that $i$ is a root of $z^4 - 2z^3 + 6z^2 - 2z + 5 = 0$, solve the equation.

Solution. The conjugate $-i$ is also a root, giving the factor $(z - i)(z + i) = z^2 + 1$. Divide (or compare coefficients) to find the other quadratic factor:

$$z^4 - 2z^3 + 6z^2 - 2z + 5 = (z^2 + 1)(z^2 + pz + q)$$

• $z^3$ terms: $p = -2$.
• Constant terms: $q = 5$.
• Check the $z^2$ terms: $q + 1 = 6$ ✓, and the $z$ terms: $p = -2$ ✓.

Now solve $z^2 - 2z + 5 = 0$:

$$z = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i$$

The four roots are $\pm i$ and $1 \pm 2i$. On an Argand diagram they’re symmetric about the real axis, as every set of roots of a real polynomial must be.

The roots of $z^4 - 2z^3 + 6z^2 - 2z + 5 = 0$ come in conjugate pairs, mirror images in the real axis.

Example 4: Finding unknown coefficients

The cubic $p(z) = z^3 + az^2 + bz + 10$, where $a, b \in \mathbb{R}$, has a root $1 + 2i$. Find the real root, and the values of $a$ and $b$.

Solution. The roots are $1 + 2i$, $1 - 2i$ and a real root $r$. Use the product of roots for a cubic ($n = 3$):

$$(1 + 2i)(1 - 2i)\,r = \frac{(-1)^3(10)}{1} = -10 \quad\Rightarrow\quad 5r = -10 \quad\Rightarrow\quad r = -2$$

Now build the polynomial from its factors:

$$p(z) = (z^2 - 2z + 5)(z + 2) = z^3 + 2z^2 - 2z^2 - 4z + 5z + 10 = z^3 + z + 10$$

So $a = 0$ and $b = 1$. The real root is $-2$.

Check: $p(-2) = -8 - 2 + 10 = 0$. ✓ And the sum of roots, $(1 + 2i) + (1 - 2i) + (-2) = 0$, matches $-a = 0$. ✓

Common mistakes

Using the theorem when the coefficients aren’t real. The conjugate root theorem only applies to polynomials with real coefficients. Check that before you write ”$a - bi$ is also a root”.

Sign errors in the quadratic factor. The pair $a \pm bi$ gives $z^2 - 2az + (a^2 + b^2)$. The middle term is minus the sum, and the constant is $a^2 + b^2$, not $a^2 - b^2$. Quick check: for $1 \pm 2i$ the factor is $z^2 - 2z + 5$, and substituting $z = 1 + 2i$ gives $(-3 + 4i) - (2 + 4i) + 5 = 0$.

Stopping too early. A cubic has three roots and a quartic has four. Once you’ve used the conjugate pair, you still need the remaining factor and its roots.

Getting the conjugate wrong. The conjugate of $-1 + 3i$ is $-1 - 3i$. Only the imaginary part changes sign.

Not checking the answer. A fast check is the sum of roots, $\dfrac{-a_{n-1}}{a_n}$. If your roots don’t add up to it, look for an arithmetic slip in the division.

Practice

1. (Warm-up) One root of a quadratic equation $z^2 + bz + c = 0$ with real coefficients is $-3 + 5i$. Write down the other root and find $b$ and $c$.

Solution

The other root is $-3 - 5i$. Sum $= -6$, product $= 9 + 25 = 34$, so the quadratic is $z^2 + 6z + 34$: $b = 6$, $c = 34$.

2. (Warm-up) Explain why a cubic equation with real coefficients cannot have the three roots $1$, $i$ and $2 + i$.

Solution

With real coefficients, the conjugates $-i$ and $2 - i$ would also have to be roots. That would make at least five roots, but a cubic has exactly three.

3. (Core) Find a cubic polynomial with real coefficients and leading coefficient $1$ that has roots $2$ and $3 - i$. Write it in expanded form.

Solution

$3 + i$ must also be a root. The pair $3 \pm i$ gives $z^2 - 6z + (9 + 1) = z^2 - 6z + 10$.

$$(z - 2)(z^2 - 6z + 10) = z^3 - 6z^2 + 10z - 2z^2 + 12z - 20 = z^3 - 8z^2 + 22z - 20$$

4. (Core) Given that $2 + i$ is a root of $z^3 - 7z^2 + 17z - 15 = 0$, find the other two roots.

Solution

$2 - i$ is also a root, giving the factor $z^2 - 4z + 5$. Comparing constant terms in $(z^2 - 4z + 5)(z + k)$: $5k = -15$, so $k = -3$.

Check: $(z^2 - 4z + 5)(z - 3) = z^3 - 3z^2 - 4z^2 + 12z + 5z - 15 = z^3 - 7z^2 + 17z - 15$. ✓

The other roots are $2 - i$ and $3$.

5. (Core) Show that $3i$ is a zero of $p(x) = x^4 - 2x^3 + 10x^2 - 18x + 9$, and find all the zeros of $p$.

Solution

Substitute, using $(3i)^2 = -9$, $(3i)^3 = -27i$ and $(3i)^4 = 81$:

$$p(3i) = 81 - 2(-27i) + 10(-9) - 18(3i) + 9 = 81 + 54i - 90 - 54i + 9 = 0$$

So $3i$ and $-3i$ are zeros, and $x^2 + 9$ is a factor. Comparing coefficients in $(x^2 + 9)(x^2 + ax + b)$: the $x^3$ terms give $a = -2$ and the constants give $9b = 9$, so $b = 1$.

$$p(x) = (x^2 + 9)(x^2 - 2x + 1) = (x^2 + 9)(x - 1)^2$$

The zeros are $3i$, $-3i$ and $1$ (a double zero).

6. (Core) The cubic $p(z) = z^3 + az^2 + bz - 20$, where $a, b \in \mathbb{R}$, has a root $2 - i$. Find the real root and the values of $a$ and $b$.

Solution

The roots are $2 - i$, $2 + i$ and a real root $r$. The product of roots is $\dfrac{(-1)^3(-20)}{1} = 20$, and $(2 - i)(2 + i) = 5$, so $5r = 20$ and $r = 4$.

$$p(z) = (z^2 - 4z + 5)(z - 4) = z^3 - 4z^2 - 4z^2 + 16z + 5z - 20 = z^3 - 8z^2 + 21z - 20$$

So $a = -8$ and $b = 21$, and the real root is $4$.

7. (Core) Find a quartic polynomial with real coefficients and leading coefficient $1$ that has roots $1 + i$ and $2 - 3i$.

Solution

The conjugates $1 - i$ and $2 + 3i$ are also roots. The quadratic factors are

$$z^2 - 2z + (1 + 1) = z^2 - 2z + 2 \qquad\text{and}\qquad z^2 - 4z + (4 + 9) = z^2 - 4z + 13$$$$\begin{aligned} (z^2 - 2z + 2)(z^2 - 4z + 13) &= z^4 - 4z^3 + 13z^2 - 2z^3 + 8z^2 - 26z + 2z^2 - 8z + 26 \\ &= z^4 - 6z^3 + 23z^2 - 34z + 26 \end{aligned}$$

8. (Challenge) Given that $z = 1 + i$ is a root of $z^3 + az + b = 0$, where $a, b \in \mathbb{R}$, find $a$ and $b$ and the other two roots.

Solution

First, $(1 + i)^2 = 2i$, so $(1 + i)^3 = 2i(1 + i) = -2 + 2i$. Substitute:

$$(-2 + 2i) + a(1 + i) + b = 0 \quad\Rightarrow\quad (-2 + a + b) + (2 + a)i = 0$$

Both parts must be zero: $2 + a = 0$ gives $a = -2$, and then $-2 - 2 + b = 0$ gives $b = 4$.

The equation is $z^3 - 2z + 4 = 0$. The roots include $1 \pm i$, giving the factor $z^2 - 2z + 2$. Comparing constants in $(z^2 - 2z + 2)(z + k)$: $2k = 4$, so $k = 2$ and the third root is $-2$.

Check: the sum of roots is $(1 + i) + (1 - i) + (-2) = 0$, which matches the missing $z^2$ term. ✓

9. (Challenge) A cubic $z^3 + az^2 + bz + c$ with real coefficients has $3i$ as a root, and the sum of its roots is $5$. Find $a$, $b$ and $c$.

Solution

The roots are $3i$, $-3i$ and a real root $r$. Their sum is $r = 5$. So

$$(z^2 + 9)(z - 5) = z^3 - 5z^2 + 9z - 45$$

and $a = -5$, $b = 9$, $c = -45$.

Check: the sum of roots is $-a = 5$ ✓, and the product is $(3i)(-3i)(5) = 45 = -c$ ✓.